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c++ - 在 Matlab 中使用来自 Opencv 的相机校准参数重新计算图像点

转载 作者:太空宇宙 更新时间:2023-11-03 22:12:29 25 4
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我已经使用OpenCv代码获得了相机标定参数,如相机矩阵、畸变系数、(旋转+平移) vector 和图像点here .

他们用来计算 2D 屏幕坐标的方程式如下: enter image description here

通过在 Matlab 中对相机矩阵、旋转和平移 vector 和物点坐标 (X,Y,Z,1) 的获得值进行硬编码,我无法获得与图像点相同的坐标值。我在这里错过了什么?我是否也需要考虑失真系数以获得准确或正确的图像点?

Matlab代码:

% Define all the parameters camera matrix , sample image point, object point,   rotation and translation vectors%
cameraMatrix = [5.9354 0 3.1950; 0 5.9354 2.3950 ; 0 0 1]
rotationMatrix = [2.5233 1.6803 3.0728];
translationMatrix = [1.2682 1.9657 8.0141];
X = [0; 0; 0; 1];

rotationMatrix = transpose(rotationMatrix);
translationMatrix = transpose(translationMatrix);

%convert the rotation vector into rotation matrix using Rodrigues func.%
rotMat = rodrigues(rotationMatrix);

R_T = horzcat(rotMat, translationMatrix)

%Convert to 2D points%
imgPts = cameraMatrix * R_T * X

lastElement = imgPts(end);

ScreenImgPts = imgPts / lastElement

对象点由棋盘正方形大小(30mm)的正方形大小定义,即 [0,0,0,1] , [30,0,0,1] 等。

但是经过我的计算和比较,存储在xml文件中的图像点是不一样的。我的结果如下

  1. 4.1343 3.8508 [0, 0, 0, 1]
  2. 3.8373 1.0331 [30, 0, 0, 1]
  3. 3.8002 0.6812 [60, 0, 0, 1]

第一个点、第二个点和第三个点的输出(图像点)应该是:

  1. 4.13546326e+002 3.85645935e+002
  2. 3.91346527e+002 3.85897003e+002
  3. 3.69121155e+002 3.86479431e+002

给出所有参数的输出文件 here

最佳答案

问题非常简单,我将所有重要参数(相机矩阵、旋转和平移)的值四舍五入到第 4 位,而指数 (e) 明显出现在值的末尾。因此四舍五入会导致不正确的值。

修改后的代码

% Define all the parameters camera matrix , sample image point, object point, rotation and translation vectors%
cameraMatrix = [5.9354136482375827e+002 0. 3.1950000000000000e+002; 0. 5.9354136482375827e+002 2.3950000000000000e+002 ; 0 0 1;]

%Rotatoin and translation vector of different planes (snapshot)%
rotationVector = [2.5233190617669338e-001 1.6802568443347082e-001 3.0727563215131681e+000];
translationVector = [1.2682348793063555e+002 1.9656574525587070e+002 8.0141048598043449e+002];

% rotationVector = [2.3492892819146791e-001 1.6451261910667694e-001 3.0787833660290516e+000];
% translationVector = [1.2806533156889765e+002 1.9877886039281353e+002 8.0447195879431570e+002];

% rotationVector = [2.1721 1.6300 3.0619];
% translationVector = [1.2661 1.9511 8.0681];

distCoeffs = [1.0829115704079707e-001 -1.0278232972256371e+000 0 0 1.7962320082487011e+000]; % k1, k2, p1, p2, k3 %

k1 = distCoeffs(1);
k2 = distCoeffs(2);
p1 = 0;
p2 = 0;
k3 = distCoeffs(end);


% X = [0 0 0; 30 0 0]

rotationVector = transpose(rotationVector);
translationVector = transpose(translationVector);


%convert the rotation vector into rotation matrix using Rodrigues func.%
rotMat = rodrigues(rotationVector)

R_T = horzcat(rotMat, translationVector)

%Convert to 2D points%
% imgPts = cameraMatrix * R_T * X
%
% lastElement = imgPts(end)
%
% ScreenImgPts = imgPts / lastElement

%%%%%%%%%%%%%% Adding calculation for distortion parameters%%%%%%%%%%

objectPoints = [0 0 0; 30 0 0; 60 0 0; 90 0 0; 120 0 0; 150 0 0;180 0 0;
0 30 0; 30 30 0; 60 30 0; 90 30 0; 120 30 0; 150 30 0; 180 30 0;
0 60 0; 30 30 0; 60 60 0; 90 60 0; 120 60 0; 150 60 0; 180 60 0;
0 90 0; 30 30 0; 60 90 0; 90 90 0; 120 90 0; 150 90 0; 180 90 0;
0 120 0; 30 120 0; 60 120 0; 90 120 0; 120 120 0; 150 120 0; 180 120 0;]

Xelement = [];
screenCoords = [];
NormXY = [];

for i = 1:35
Xelement = (objectPoints(i,:))
NormXY(:,1) = (rotMat * transpose(Xelement)) + translationVector
lastElement = NormXY(end)
NormXY = NormXY / lastElement
x = NormXY(1)
y = NormXY(2)

r2 = power(x,2) + power(y,2)
r4 = power(r2,2)
r6 = power(r2,3)

xcorr = x * (1 + k1*r2 + k2*r4 + k3*r6)
ycorr = y * (1 + k1*r2 + k2*r4 + k3*r6)

XY = [xcorr ;ycorr;1]
screenCoords(:,i) = cameraMatrix * XY


end

关于c++ - 在 Matlab 中使用来自 Opencv 的相机校准参数重新计算图像点,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/31772459/

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